To find: The value of .
The value of is .
The company makes square wafers of silicon of area of the form .
Differentiate the area of the square wafer respect to x.
Substitute 15 for in the above equation and find as,
Therefore, the rate at which the area will increase with respect to change in its side length, if the side length is 15 mm will be .
To Show: The rate of change of the area of a square with respect to the change in its side length is equal to half its perimeter.
Draw the square with side length, x as shown in Figure (1).
The perimeter of the square is .
From the figure 1, the change in length is very small.
The first derivative of is,
Thus, the change in area of the square is approximately half of its perimeter which is half of the 4 sides times.
The expression for the shaded area (change in area) as below.
The change in length is small, and change in area is approximately equal to .
The term is approximately equal to .
Thus the rate of change of the area of a square with respect to the change in its side length is equal to half its perimeter is .
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!